Comparison of reaching probabilities of two Markov chains

Question

Let $N_n:=\{1,2,\cdots,n\}$. Given two finite states Markov chains $\big(X^{(j)}_i\in N_n\}\big)_{i=0}^\infty$ for $j\in\{1,2\}$. Pr$(X^{(1)}_{i+1}=b|X_i=a)>$Pr$(X^{(2)}_{i+1}=b|X_i=a), \,\forall a<b, a,b\in N_n$. Pr$(X^{(1)}_{i+1}=b|X_i=a)\le $Pr$(X^{(2)}_{i+1}=b|X_i=a), \,\forall a\ge b, a,b\in N_n$. Is it true that Pr$(X^{(1)}$ reaches $b|X^{(1)}_0=a)>$Pr$(X^{(2)}$ reaches $b|X^{(2)}_0=a), \,\forall a<b$, and Pr$(X^{(1)}$ reaches $b|X^{(1)}_0=a)\le$Pr$(X^{(2)}$ reaches $b|X^{(2)}_0=a), \,\forall a \ge b$?

Does a coupling argument help to resolve this problem?

Answer 1

So the way you have set it up, $X^{(1)}$ moves to the right more consistently than $X^{(2)}$, and $X^{(2)}$ moves to the left more consistently than $X^{(1)}$, in the sense that if their transition matrices are $P$ and $Q$ respectively then $p_{ij}>q_{ij}$ if $i<j$ and $p_{ij} \leq q_{ij}$ if $i \geq j$.

So consider:

• $q_{11}=1$, $q_{ij}>0$ if $i \neq 1$
• $p_{ij}>0$ for all $i,j$.
• Otherwise require the above inequalities.

Now $X^{(2)}$ may fail to visit, say, $2$, even if it begins at a state to the right of $2$, because it could potentially skip directly to $1$ and then be stuck there forever. Meanwhile $X^{(1)}$ will visit all states with probability $1$.